Bar Magnet and Magnetic Dipole — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Bar Magnet and Magnetic Dipole MCQs with step-by-step solutions (9 questions). Part of Magnetism and Matter. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Bar Magnet and Magnetic Dipole · medium · theory
A uniform conducting wire of length $12a$ and resistance $R$ is wound up as a current ($I$) carrying coil in the shape of (1) an equilateral triangle of side $a$, and (2) a square of side $a$. The magnetic dipole moments of the coil in each case respectively are
A. $\sqrt{3}\,Ia^2$ and $3Ia^2$ ✓ Correct
B. $3Ia^2$ and $Ia^2$
C. $3Ia^2$ and $4Ia^2$
D. $4Ia^2$ and $3Ia^2$
Solution: Triangle: perimeter $3a\Rightarrow n=4$ turns, area $=\dfrac{\sqrt3}{4}a^2$, so $M=nIA=4I\cdot\dfrac{\sqrt3}{4}a^2=\sqrt3\,Ia^2$. Square: perimeter $4a\Rightarrow n=3$ turns, area $=a^2$, so $M=nIA=3Ia^2$.
Q2 — Bar Magnet and Magnetic Dipole · medium · theory
A wire of length $L$ metre carrying a current of $I$ ampere is bent in the form of a circle. Its magnetic moment is
A. $\dfrac{IL^2}{4}\,\text{Am}^2$
B. $\dfrac{\pi IL^2}{4}\,\text{Am}^2$
C. $\dfrac{2IL^2}{\pi}\,\text{Am}^2$
D. $\dfrac{IL^2}{4\pi}\,\text{Am}^2$ ✓ Correct
Solution: Circumference $L=2\pi r\Rightarrow r=\dfrac{L}{2\pi}$. Area $A=\pi r^2=\dfrac{L^2}{4\pi}$. Magnetic moment $M=IA=\dfrac{IL^2}{4\pi}\,\text{Am}^2$.
Q3 — Bar Magnet and Magnetic Dipole · hard · numerical
A 250-turn rectangular coil of length $2.1\,\text{cm}$ and width $1.25\,\text{cm}$ carries a current of $85\,\mu\text{A}$ and is subjected to a magnetic field of strength $0.85\,\text{T}$. The work done for rotating the coil by $180^\circ$ against the torque is
A. $9.1\,\mu\text{J}$ ✓ Correct
B. $4.55\,\mu\text{J}$
C. $2.3\,\mu\text{J}$
D. $1.5\,\mu\text{J}$
Solution: $W=2MB=2NIAB=2\times250\times(85\times10^{-6})\times(2.1\times1.25\times10^{-4})\times0.85\approx9.5\times10^{-6}\,\text{J}$, closest to $9.1\,\mu\text{J}$.
Q4 — Bar Magnet and Magnetic Dipole · medium · theory
A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by $60^\circ$ is $W$. Now the torque required to keep the magnet in this new position is
A. $\dfrac{W}{\sqrt3}$
B. $\sqrt3\,W$ ✓ Correct
C. $\dfrac{\sqrt3\,W}{2}$
D. $\dfrac{2W}{\sqrt3}$
Solution: $W=MB(\cos0^\circ-\cos60^\circ)=MB\left(1-\dfrac12\right)=\dfrac{MB}{2}\Rightarrow MB=2W$. Torque at $60^\circ$ is $\tau=MB\sin60^\circ=2W\cdot\dfrac{\sqrt3}{2}=\sqrt3\,W$.
Q5 — Bar Magnet and Magnetic Dipole · medium · theory
A bar magnet of length $l$ and magnetic dipole moment $M$ is bent in the form of a circular arc that subtends an angle of $60^\circ$ at the centre of the circle (i.e. the magnet's length equals one-sixth of the circle's circumference). The new magnetic dipole moment will be
A. $M$
B. $\dfrac{3}{\pi}M$ ✓ Correct
C. $\dfrac{2}{\pi}M$
D. $\dfrac{M}{2}$
Solution: Since the arc subtends $60^\circ$, $l=\dfrac{2\pi r}{6}=\dfrac{\pi r}{3}\Rightarrow r=\dfrac{3l}{\pi}$. Pole strength $m=\dfrac{M}{l}$, and the new moment (pole strength times the straight-line distance between poles) is $M'=m\cdot r=\dfrac{M}{l}\cdot\dfrac{3l}{\pi}=\dfrac{3M}{\pi}$.
Q6 — Bar Magnet and Magnetic Dipole · easy · theory
A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It
A. will become rigid showing no movement
B. will stay in any position ✓ Correct
C. will stay in North-South direction only
D. will stay in East-West direction only
Solution: At a geomagnetic pole the horizontal component of the earth's magnetic field is zero, so there is no restoring torque on the needle and it can stay in any position.
Q7 — Bar Magnet and Magnetic Dipole · easy · numerical
A bar magnet having a magnetic moment of $2\times10^{4}\,\text{J T}^{-1}$ is free to rotate in a horizontal plane. A horizontal magnetic field $B=6\times10^{-4}\,\text{T}$ exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60^\circ$ from the field is
A. $0.6\,\text{J}$
B. $12\,\text{J}$
C. $6\,\text{J}$ ✓ Correct
D. $2\,\text{J}$
Solution: $W=MB(\cos0^\circ-\cos60^\circ)=2\times10^{4}\times6\times10^{-4}\times\left(1-\dfrac12\right)=6\,\text{J}$.
Q8 — Bar Magnet and Magnetic Dipole · medium · theory
A charged particle (charge $q$) is moving in a circle of radius $R$ with uniform speed $v$. The associated magnetic moment $\mu$ is given by
A. $\dfrac{qvR}{2}$ ✓ Correct
B. $qvR^2$
C. $\dfrac{qvR^2}{2}$
D. $qvR$
Solution: Current equivalent to the revolving charge is $i=\dfrac{qv}{2\pi R}$, so $\mu=iA=\dfrac{qv}{2\pi R}\times\pi R^2=\dfrac{qvR}{2}$.
Q9 — Bar Magnet and Magnetic Dipole · easy · theory
A bar magnet of magnetic moment $\vec{M}$ is placed in a magnetic field of induction $\vec{B}$. The torque exerted on it is
A. $\vec{M}\cdot\vec{B}$
B. $-\vec{M}\cdot\vec{B}$
C. $\vec{M}\times\vec{B}$ ✓ Correct
D. $-\vec{M}\times\vec{B}$
Solution: The torque on a magnetic dipole placed in an external field is $\vec\tau=\vec{M}\times\vec{B}$.